的士司机接客游戏简介：
在这个游戏中，黄色方块代表出租车，(“|”)表示一堵墙，蓝色字母代表接乘客的位置，紫色字母是乘客下车的位置，出租车上有乘客时就会变绿。你作为的士司机要找到最快速接客和下客的路径。
接到客人了
放下客人，完成！
思路：
客人和下客的位置是随机出现的，不可能用机械的方法找出变动的路径，可以用人工智能强化学习方法学习出最优路径。
代码：
import gym
import numpy as np
env = gym.make('Taxi-v2')
Q = np.zeros((env.observation_space.n,env.action_space.n))
def trainQ():
for _ in range(10000):
observation = env.reset()
while True:
action = env.action_space.sample()
observation_,reward, done,info = env.step(action)
Q[observation,action] = reward + 0.75 * Q[observation_].max()
observation = observation_
if done:break
return Q
def findway():
observation = env.reset()
rewards = 0
while True:
action = Q[observation].argmax()
observation_,reward, done,info = env.step(action)
print(observation_,reward, done,info)
rewards += reward
observation = observation_
env.render()
if done:
print(rewards)
break
Q = trainQ()
findway()
测试：
客人在这
接到客人了
开车
开车
开车
下客，完工！
倒立摆游戏简介：
思路：
倒立摆游戏比的士司机游戏复杂，原因在于倒立摆的连续状态是无穷多个，人工智能 Q-learning 方法需要有限个状态形成知识。
解决方案：只需要将连续状态打散为离散状态即可。
代码：
import gym
import numpy as np
env = gym.make('CartPole-v0')
eplision = 0.01
q_table = np.zeros((256,2))
def bins(clip_min, clip_max, num):
return np.linspace(clip_min, clip_max, num + 1)[1:-1]
def digitize_state(observation):
cart_pos, cart_v, pole_angle, pole_v = observation
digitized = [np.digitize(cart_pos, bins=bins(-2.4, 2.4, 4)),
np.digitize(cart_v, bins=bins(-3.0, 3.0, 4)),
np.digitize(pole_angle, bins=bins(-0.5, 0.5, 4)),
np.digitize(pole_v, bins=bins(-2.0, 2.0, 4))]
return sum([x * (4 ** i) for i, x in enumerate(digitized)])
#------------观察期--------------#
for _ in range(1000):
observation = env.reset()
s = digitize_state(observation)
while True:
action = env.action_space.sample()
observation_, reward, done, info = env.step(action)
if done:reward = -200
s_ = digitize_state(observation_)
action_ = env.action_space.sample()
q_table[s,action] = reward + 0.85*q_table[s_,action_]
s,action = s_,action_
if done:break
print('观察期结束')
#------------贪心策略期--------------#
for epicode in range(1000):
observation = env.reset()
s = digitize_state(observation)
while True:
eplision = epicode / 1000
action = q_table[s].argmax() if np.random.random() < eplision else env.action_space.sample()
observation_, reward, done, info = env.step(action)
if done:reward = -200
s_ = digitize_state(observation_)
action_ = q_table[s_].argmax() if np.random.random() < eplision else env.action_space.sample()
q_table[s,action] = reward + 0.85*q_table[s_,action_]
s,action = s_ ,action_
if done:break
print('贪心策略期结束')
#------------验证期--------------#
scores = []
for _ in range(100):
score = 0
observation = env.reset()
s = digitize_state(observation)
while True:
action = q_table[s].argmax()
observation_, reward, done, info = env.step(action)
score += reward
s = digitize_state(observation_)
#env.render()
if done:
scores.append(score)
break
print('验证期结束\n验证成绩：%s'%np.max(scores))
测试：
观察期结束
贪心策略期结束
验证期结束
验证成绩：200.0
倒立摆
纪要：
离散化函数原先为
def digitize_state(observation,bin=5):
high_low = np.vstack([env.observation_space.high,env.observation_space.low]).T[:,::-1]
bins = np.vstack([np.linspace(*i,bin) for i in high_low])
state = [np.digitize(state,bin).tolist() for state,bin in zip(observation,bins)]
state = sum([value*2**index for index,value in enumerate(state)])
return state
效果并不好，在网上找了个好的替换：
def bins(clip_min, clip_max, num):
return np.linspace(clip_min, clip_max, num + 1)[1:-1]
def digitize_state(observation):
cart_pos, cart_v, pole_angle, pole_v = observation
digitized = [np.digitize(cart_pos, bins=bins(-2.4, 2.4, 4)),
np.digitize(cart_v, bins=bins(-3.0, 3.0, 4)),
np.digitize(pole_angle, bins=bins(-0.5, 0.5, 4)),
np.digitize(pole_v, bins=bins(-2.0, 2.0, 4))]
return sum([x * (4 ** i) for i, x in enumerate(digitized)])
可见要做好人工智能还是需要了解状态参数大致意义，不管含义什么都喂进智能体的话，智能体表现不会太好。
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